Answer
$f'(x) = -\frac{12}{(2x+1)^{3}}$
Work Step by Step
The result of Exercise $75$ says that given $f(x) = (mx+b)^{n}$, then $f'(x) = nm(mx+b)^{n-1}$
Applying this result to $f(x) = \frac{3}{(2x+1)^2} = 3(2x+1)^{-2}$ where $m=2$, $n=-2$, and $b=1$, we get:
$f'(x) = 3(\frac{d}{dx} \frac{1}{(2x+1)^2}) = 3(nm(mx+b)^{n-1})= (-2)(2)(2x+1)^{((-2)-1)} =3(-4(2x+1)^{-3})= -\frac{12}{(2x+1)^{3}}$
